3. 118
printf("%.5ft",a[i][j]);
printf("n");
}
printf("n");
t=1;
flushall();
while (t)
{
printf("Co sua ma tran a khong(c/k)?");
scanf("%c",&tl);
if (toupper(tl)=='C')
{
printf("Cho chi so hang can sua : ");
scanf("%d",&i);
printf("Cho chi so cot can sua : ");
scanf("%d",&j);
printf("a[%d][%d] = ",i,j);
scanf("%f",&a[i,j]);
}
if (toupper(tl)=='K')
t=0;
}
printf("Ma tran a ban daun");
printf("n");
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
printf("%.5ft",a[i][j]);
printf("n");
}
printf("n");
d=1;
i=1;
ok2=1;
while ((ok2)&&(i<=n))
{
if (a[i][i]==0)
{
ok1=1;
k=k+1;
while ((ok1)&&(k<=n))
if (a[k,i]!=0)
{
for (j=i;j<=n;j++)
{
c=a[i][j];
a[i][j]=a[k][j];
a[k][j]=c;
}
d=-d;
4. 119
ok1=0;
}
else
k=k+1;
if (k>n)
{
printf("n");
printf("** MA TRAN SUY BIEN **");
ok2=0;
d=0;
}
}
if (a[i][i]!=0)
{
c=a[i][i];
for (j=i+1;j<=n;j++)
a[i][j]=a[i][j]/c;
for (k=i+1;k<=n;k++)
{
c=a[k][i];
for (j=i+1;j<=n;j++)
a[k][j]=a[k][j]-a[i][j]*c;
}
}
i=i+1;
}
if (ok2)
{
for (i=1;i<=n;i++)
d=d*a[i][i];
printf("n");
printf("** GIA TRI DINH THUC D **");
printf("n");
printf("%.3f",d);
}
getch();
}
§2.NghÞch ®¶o ma trËn
Gäi A-1
lµ ma trËn nghÞch ®¶o cña mét ma trËn A bËc n ta cã AA-1
= E.(trong biÓu
thøc nµy E lµ mét ma trËn vu«ng cã c¸c phÇn tö trªn ®−êng chÐo chÝnh b»ng 1). D¹ng cña
ma trËn E,vÝ dô cÊp 4,lµ:
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
=
1000
0100
0010
0001
E
6. 121
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−−
−−
−−
=
−
434141
414341
414143
A
1
¸p dông ph−¬ng ph¸p nµy chóng ta cã ch−¬ng tr×nh sau:
Ch−¬ng tr×nh 9-2
#include <conio.h>#include <stdio.h>#include <math.h>#include <stdlib.h>#include
<ctype.h>void main() { int i,j,k,n,t,t1;float c,a[50][50],b[50][50]; char tl; clrscr();
printf(" **MA TRAN NGHICH DAO** n");
printf("Cho bac cua ma tran n = ");
scanf("%d",&n);
printf("Vao ma tran ban dau an");
for (i=1;i<=n;i++)
{
printf("Vao hang thu %d :n",i);
for (j=1;j<=n;j++)
{
printf("a[%d][%d] = ",i,j);
scanf("%f",&a[i][j]);
}
printf("n");
}
printf("n");
printf("Ma tran ban da nhapn");
printf("n");
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
printf("%.5ft",a[i][j]);
printf("n");
}
t=1;
flushall();
while (t)
{
printf("nCo sua ma tran khong(c/k)?");
scanf("%c",&tl);
if(toupper(tl)=='C')
{
printf("Cho chi so hang can sua : ");
scanf("%d",&i);
printf("Cho chi so cot can sua : ");
scanf("%d",&j);
printf("a[%d][%d] = ",i,j);
scanf("%f",&a[i][j]);
}
if (toupper(tl)=='K')
t=0;
}
printf("nMa tran ban daun");
7. 122
printf("n");
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
printf("%.5ft",a[i][j]);
printf("n");
}
printf("n");
for (i=1;i<=n;i++)
for (j=n+1;j<=2*n;j++)
{
if (j==i+n)
a[i][j]=1;
else
a[i][j]=0;
}
i=1;
t1=1;
while (t1&&(i<=n))
{
if (a[i][i]==0)
{
t=1;
k=i+1;
while (t&&(k<=n))
if (a[k][i]!=0)
{
for (j=1;j<=2*n;j++)
{
c=a[i][j];
a[i][j]=a[k][j];
a[k][j]=c;
}
t=0;
}
else
k=k+1;
if (k==n+1)
{
if (a[i][k-1]==0)
{
printf("MA TRAN SUY BIENn ");
t1=0;
}
}
}
if (a[i][i]!=0)
{
c=a[i][i];
for (j=i;j<=2*n;j++)
9. 124
{
int n,l,m,i,j,k,t;
float a[max][max],b[max][max],c[max][max];
char tl;
clrscr();
printf("Cho so hang cua ma tran a : ");
scanf("%d",&n);
printf("Cho so cot cua ma tran a : ");
scanf("%d",&l);
printf("Cho so cot cua ma tran b : ");
scanf("%d",&m);
printf("nNHAP MA TRAN An");
for (i=1;i<=n;i++)
for (j=1;j<=l;j++)
{
printf("a[%d][%d] = ",i,j);
scanf("%f",&a[i][j]);
}
printf("n");
printf("Ma tran a ma ban da nhapn");
for (i=1;i<=n;i++)
{
for (j=1;j<=l;j++)
printf("%10.5f",a[i][j]);
printf("n");
}
flushall();
t=1;
while (t)
{
printf("Co sua ma tran khong(c/k)?");
scanf("%c",&tl);
if (toupper(tl)=='C')
{
printf("Cho chi so hang can sua : ");
scanf("%d",&i);
printf("Cho chi so cot can sua : ");
scanf("%d",&j);
printf("a[%d][%d] = ",i,j);
scanf("%f",&a[i][j]);
}
if (toupper(tl)=='K')
t=0;
}
printf("Ma tran a ban dau");
printf("n");
for (i=1;i<=n;i++)
{
for (j=1;j<=l;j++)
10. 125
printf("%10.5f",a[i][j]);
printf("n");
}
printf("n");
printf("NHAP MA TRAN Bn");
for (i=1;i<=l;i++)
for (j=1;j<=m;j++)
{
printf("b[%d][%d] = ",i,j);
scanf("%f",&b[i][j]);
}
printf("n");
printf("Ma tran b ban da nhapn");
for (i=1;i<=l;i++)
{
for (j=1;j<=m;j++)
printf("%10.5f",b[i][j]);
printf("n");
}
flushall();
t=1;
while (t)
{
printf("Co sua ma tran khong(c/k)?");
scanf("%c",&tl);
if (toupper(tl)=='C')
{
printf("Cho chi so hang can sua : ");
scanf("%d",&i);
printf("Cho chi so cot can sua : ");
scanf("%d",&j);
printf("b[%d][%d] = ",i,j);
scanf("%f",&b[i][j]);
}
if (toupper(tl)=='K')
t=0;
}
printf("Ma tran b ban dau");
printf("n");
for (i=1;i<=l;i++)
{
for (j=1;j<=m;j++)
printf("%10.5f",b[i][j]);
printf("n");
}
printf("n");
for (i=1;i<=n;i++)
for (j=1;j<=m;j++)
{
11. 126
c[i][j]=0;
for (k=1;k<=l;k++)
c[i][j]=c[i][j]+a[i][k]*b[k][j];
}
printf("Ma tran tich c :n");
for (i=1;i<=n;i++)
{
for (j=1;j<=m;j++)
printf("%10.5f",c[i][j]);
printf("n");
}
getch();
}
Dïng ch−¬ng tr×nh tÝnh tÝnh hai ma trËn ta nhËn ®−îc kÕt qu¶
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
−−
−
−
−
=⎟
⎠
⎞
⎜
⎝
⎛
−
−×
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
−
1214
221
11148
105
343
221
35
01
31
12
§4.Gi¸ trÞ riªng vµ vec t¬ riªng cña ma trËn
1.Kh¸i niÖm chung: Trong nghiªn lÝ thuyÕt vµ øng dông,ta gÆp bµi to¸n vÒ ma trËn cÊp
n.Cho mét ma trËn A cÊp n,gi¸ trÞ λ ®−îc gäi lµ gi¸ trÞ riªng vµ vect¬ X ®−îc gäi lµ vect¬
riªng cña ma trËn A nÕu:
AX = λX (1)
Vect¬ riªng ph¶i lµ vect¬ kh¸c kh«ng.T−¬ng øng víi mét gi¸ trÞ riªng cã v« sè vect¬
riªng.NÕu X lµ mét vÐc t¬ riªng t−¬ng øng víi gi¸ trÞ riªng λ th× cX còng lµ vec t− riªnh øng
víi λ.Cã nhiÒu thuËt to¸n t×m gi¸ trÞ riªng vµ vect¬ riªng cña mét ma trËn.Gi¶ sö ta cã ma
trËn A,gäi E lµ ma trËn ®¬n vÞ th× theo (1) ta cã:
(A-λE)X = 0 (2)
vµ (A - λE) lµ ma trËn cã d¹ng:
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎝
⎛
−
−
−
λ
λ
λ
aaa
.......
aaa
aaa
nn2n1n
n22221
n11211
....
....
....
(3)
Nh− vËy do (2) lµ hÖ ph−¬ng tr×nh tuyÕn tÝnh thuÇn nhÊt nªn ®iÒu kiÖn cÇn vµ ®ñ ®Ó λ
lµ gi¸ trÞ riªng cña ma trËn trªn lµ ®Þnh thøc cña nã b»ng kh«ng:
det(A - λE) = 0 (4)
Ph−¬ng tr×nh (4) ®−îc gäi lµ ph−¬ng tr×nh ®Æc tr−ng cña ma trËn A.§Þnh thøc det(A - λE)
®−îc gäi lµ ®Þnh thøc ®Æc tr−ng cña ma trËn A.§Þnh thøc PA(λ) cña ma trËn trªn ®−îc gäi lµ
®a thøc ®Æc tr−ng cña ma trËn vu«ng A.
VÝ dô t×m vec t¬ riªng vµ trÞ riªng cña ma trËn:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
022
113
313
Tr−íc hÕt ta tÝnh ®a thøc ®Æc tr−ng cña ma trËn A:
13. 128
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
{
printf("a[%d][%d] = ",i,j );
scanf("%f",&a[i][j]);
}
printf("n");
clrscr();
printf("Ma tran ban da nhap");
printf("n");
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
printf("%10.5f",a[i][j]);
printf("n");
}
t=1;
flushall();
while (t)
{
printf("n");
printf("Co sua ma tran khong(c/k)?");
scanf("%c",&tl);
if (toupper(tl)=='C')
{
printf("Cho chi so hang can sua : ");
scanf("%d",&i);
printf("Cho chi so cot can sua : ");
scanf("%d",&j);
printf("a[%d][%d] = ",i,j);
scanf("%f",&a[i][j]);
flushall();
}
if (toupper(tl)=='K')
t=0;
}
printf("Ma tran ban dau");
printf("n");
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
printf("%10.5f",a[i][j]);
printf("n");
}
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
b[i][j]=a[i][j];
for (k=1;k<=n-1;k++)
{
vet=0.0;
16. 131
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
−−−−
=
26544323
68102
720138
17302417
A
Chän V= {1,1,1,1}T
ta tÝnh ®−îc
V V1 = AV V’1 V2 =
AV’1
V’2
1 88 -0.6027 -6.4801 -0.5578
1 48 -0.3288 -5.6580 -0.4870
1 26 -0.1781 0.0818 0.0070
1 -146 1 11.6179 1
λ 11.6179
V3 =
AV’2
V’3 V4 = AV’3 V’4 V5 =
AV’4
-3.9594 -0.5358 -3.6823 -0.5218 -3.5718
-3.6526 -0.4942 -3.5196 -0.4987 -3.4791
0.0707 0.0096 0.0630 0.0089 0.0408
7.3902 1 7.0573 1 6.9638
λ 7.3902 7.0573 6.9638
V’5 V6=
AV’5
V’6 V7= AV’6 V’7
-
0.5129
-3.5341 -0.5075 -3.5173 -0.5043
-
0.4996
-3.4809 -0.4999 -3.4868 -0.5000
0.0059 0.0250 0.0036 0.0147 0.0021
1 6.9634 1 6.9742 1
λ 6.9634 6.9742
Dïng thuËt to¸n trªn ta cã ch−¬ng tr×nh sau:
Ch−¬ng tr×nh 9-5
#include <conio.h>#include <stdio.h>#include <math.h>#include <stdlib.h>#include
<ctype.h>#define max 50void main() { int i,j,k,n,t; char tl; float t0,t1,epsi,s;
float a[max][max];
float x0[max],x1[max];
clrscr();
printf("Phuong phap lap luy thua tim tri rieng lon nhatn");
printf("Cho so hang va cot cua ma tran n = ");
scanf("%d",&n);
printf("Cho cac phan tu cua ma tran a : n");
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
17. 132
{
printf("a[%d][%d] = ",i,j);
scanf("%f",&a[i][j]);
}
printf("n");
printf("Ma tran ban da nhapn");
printf("n");
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
printf("%15.5f",a[i][j]);
printf("n");
}
flushall();
t=1;
while (t)
{
printf("nCo sua ma tran khong(c/k)?");
scanf("%c",&tl);
if (toupper(tl)=='C')
{
printf("Cho chi so hang can sua : ");
scanf("%d",&i);
printf("Cho chi so cot can sua : ");
scanf("%d",&j);
printf("a[%d][%d] = ",i,j);
scanf("%f",&a[i][j]);
}
if (toupper(tl)=='K')
t=0;
}
epsi=1e-5;
printf("nMa tran ban daun");
printf("n");
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
printf("%15.5f",a[i][j]);
printf("n");
}
printf("n");
for (i=1;i<=n;i++)
x0[i]=1;
k=1;
t=0;
t1=0;
do
{
t0=t1;
for (i=1;i<=n;i++)
20. 135
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
−−−−
=
26544323
68102
720138
17302417
A
Ta ®· t×m ®−îc gi¸ trÞ riªng lín nhÊt λ1 = 7 vµ mét vect¬ riªng t−¬ng øng:
X1 = { 1,1,0,-2}T
.
Ma trËn AT
cã d¹ng:
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
−
−
−
−
=
266717
5482030
43101324
232817
A
T
vµ theo ph−¬ng tr×nh AT
-λ1E)W1 = 0 ta t×m ®−îc vect¬ W1 = {293,695,746,434}T
Ta lËp ma trËn míi A1 theo (7):
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
−−−−
=λ
86814921390586
0000
434746695293
434746695293
120
7
XW
WX
1
T
1
T
11
1
vµ:
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
−−−−
−−−−
=
6333.240333.330833.381833.11
68102
3167.185167.235417.270917.9
3167.85167.135417.160917.0
A1
Tõ ma trËn A1 ta t×m tiÕp ®−îc λ2 theo phÐp lÆp luü thõa vµ sau ®ã l¹i t×m ma trËn A3 vµ t×m
gi¸ trÞ riªng t−¬ng øng.
Ch−¬ng tr×nh lÆp t×m c¸c gi¸ trÞ riªng vµ vec t¬ riªng cña ma trËn nh− sau:
Ch−¬ng tr×nh 9-6
#include <conio.h>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <ctype.h>
#define max 50
void main()
{
float a[max][max],vv[max][max],at[max][max];
float x[max],y[max],vd[max];
int i,j,k,n,l,t;
float vp,v1,z,epsi,va,ps;
char tl;
clrscr();
epsi=0.000001;
printf("Cho bac cua ma tran n = ");
scanf("%d",&n);
printf("Cho cac phan tu cua ma tran a : n");
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
21. 136
{
printf("a[%d][%d] = ",i,j);
scanf("%f",&a[i][j]);
}
printf("n");
clrscr();
printf("Ma tran ban da nhap");
printf("n");
for (i=1;i<=n;i++)
{
for (j=1;j<=n;j++)
printf("%15.5f",a[i][j]);
printf("n");
}
t=1;
flushall();
while (t)
{
printf("n");
printf("Co sua ma tran khong(c/k)?");
scanf("%c",&tl);
if (toupper(tl)=='C')
{
printf("Cho chi so hang can sua : ");
scanf("%d",&i);
printf("Cho chi so cot can sua : ");
scanf("%d",&j);
printf("a[%d][%d] = ",i,j);
scanf("%f",&a[i][j]);
}
if (toupper(tl)=='K')
t=0;
}
for (l=1;l<=n;l++)
{
for (i=1;i<=n;i++)
x[i]=1;
vp=1.23456789;
k=0;
for (k=1;k<=40;k++)
{
for (i=1;i<=n;i++)
{
y[i]=0;
for (j=1;j<=n;j++)
y[i]=y[i]+a[i][j]*x[j];
}
v1=y[1]/x[1];
z=0;
for (i=1;i<=n;i++)
22. 137
if (fabs(y[i])>z)
z=y[i];
for (i=1;i<=n;i++)
x[i]=y[i]/z;
if (fabs(vp-v1)<epsi)
break;
vp=v1;
}
{
printf("Gia tri rieng : %9.6fn",v1);
printf("Vec to rieng : n");
for (i=1;i<=n;i++)
printf("%.5fn",x[i]);
printf("n");
getch();
}
vd[l]=v1;
va=v1;
for (i=1;i<=n;i++)
vv[l][i]=x[i];
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
at[i][j]=a[j][i];
for (i=1;i<=n;i++)
x[i]=1;
vp=1.23456;
k=0;
for (k=1;k<=40;k++)
{
for (i=1;i<=n;i++)
{
y[i]=0;
for (j=1;j<=n;j++)
y[i]=y[i]+at[i][j]*x[j];
}
v1=y[1]/x[1];
z=0;
for (i=1;i<=n;i++)
if (fabs(y[i])>z)
z=y[i];
for (i=1;i<=n;i++)
x[i]=y[i]/z;
if (fabs(vp-v1)<epsi)
break;
vp=v1;
}
if (fabs(vp-v1)>epsi)
{
printf("Khong hoi tu sau 40 lan lapn");
getch();
27. 142
J(Xi)(Xi+1 - Xi) = -F(Xi)
víi Xi = { si,pi}T
vµ Xi+1 = { si+1,pi+1}T
)p,s(g
)p,s(f
)X(F
ii
ii
i =
i
J X
f
s
f
p
g
s
g
p
( )=
∂
∂
∂
∂
∂
∂
∂
∂
Quan hÖ : J(Xi)∆X = -F(Xi) víi ∆X = {si+1 - si,pi+1 - pi}T
t-¬ng øng víi mét hÖ ph-¬ng tr×nh
tuyÕn tÝnh hai Èn sè ∆s = si+1 - si vµ ∆p = pi+1 - pi :
∂
∂
∂
∂
∂
∂
∂
∂
f
s
s
f
p
p f s p
g
s
s
g
p
p g s p
i i
i i
∆ ∆
∆ ∆
+ =
+ =
⎧
⎨
⎪
⎪
⎪
⎩
⎪
⎪
⎪
−
−
( )
( )
,
,
Theo c«ng thøc Cramer ta cã :
∆s
f
g
p
g
f
p
=
− +
∂
∂
∂
∂
δ
∆p
g
f
s
f
g
s=
− +
∂
∂
∂
∂
δ
δ
∂
∂
∂
∂
∂
∂
∂
∂
= −
f
s
g
p
f
p
g
s
§Ó dïng ®-îc c«ng thøc nµy ta cÇn tÝnh ®-îc c¸c ®¹o hµm
∂
∂
f
s
,
∂
∂
f
p
,
∂
∂
g
s
,
∂
∂
g
p
.C¸c ®¹o hµm
nµy ®-îc tÝnh theo c«ng thøc truy håi.
Do bo = ao nªn
ob
s
∂
∂
=0 ob
p
∂
∂
=0
b1 = a1 + sbo nªn 1∂
∂
b
s
bo
= 1
0
∂
∂
b
p
=
b2 = a2 + sb1- pbo nªn 2 2 1∂
∂
∂
∂
∂
∂
∂
∂
b
s
a
s
sb
s
pb
s
o
= + −
( ) ( )
MÆt kh¸c : 2
0
∂
∂
a
s
=
1 1
1
∂
∂
∂
∂
( )sb
s s
b
s b= +
o
pb
s
∂
∂
( )
=0
nªn :
2
1
∂
∂
b
s
b sbo
= +
b3 = a3 + sb2- pb1 nªn
3
2
2 1∂
∂
∂
∂
∂
∂
b
s
b s
b
s
p b
s
= + −
NÕu chóng ta ®Æt :
k
k
b
s c
∂
∂
= −1
33. 148
Ch−¬ng tr×nh 8-11
//giai he pt phi tuyen
#include <conio.h>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#define n 4
float a[n+1][n+2];
float x[n+1],y[n+1];
int i,j,k,l,z,r;
float e,s,t;
void main()
{
void doc();
clrscr();
printf("Cho cac gia tri nghiem ban daun");
for (i=1;i<=n;i++)
{
printf("x[%d] = ",i);
scanf("%f",&x[i]);
}
e=1e-6;
z=30;
for (r=1;r<=z;r++)
{
doc();
for (k=1;k<=n-1;k++)
{
s=0 ;
for (i=k;i<=n;i++)
{
t=fabs(a[i][k]);
if (s<=t)
{
s=t;
l=i;
}
}
for (j=k;j<=n+1;j++)
{
s=a[k][j];
a[k][j]=a[l][j];
a[l][j]=s;
}
if (a[1][1]==0)
{
34. 149
printf("Cac phan tu duong cheo cua ma tran bang khong");
getch();
exit(1);
}
else
{
if (fabs(a[k][k]/a[1][1])<(1e-08))
{
printf("Ma tran suy bien");
goto mot;
}
}
for (i=k+1;i<=n;i++)
{
if (a[k][k]==0)
{
printf("Cac phan tu duong cheo cua ma tran bang
khongn");
goto mot;
}
s=a[i][k]/a[k][k];
a[i][k]=0;
for (j=k+1;j<=n+1;j++)
a[i][j]=a[i][j]-s*a[k][j];
}
y[n]=a[n][n+1]/a[n][n];
for (i=n-1;i>=1;i--)
{
s=a[i][n+1];
for (j=i+1;j<=n;j++)
s=s-a[i][j]*y[j];
if (a[i][i]==0)
{
printf("Cac phan tu duong cheo cua ma tran bang
khongn");
goto mot;
}
y[i]=s/a[i][i];
}
}
if (r!=1)
for (i=1;i<=n;i++)
{
if (fabs(y[i])<e*fabs(x[i]))
goto ba;
}
for (i=1;i<=n;i++)
x[i]=x[i]-y[i];
printf("n");
}
35. 150
printf("Khong hoi tu sau %d lan lapn",z);
goto mot;
clrscr();
ba:printf("Vec to nghiemn");
for (i=1;i<=n;i++)
printf("%.5fn",(x[i]-y[i]));
printf("n");
printf("Do chinh xac cua nghiem la %.5f: n", e);
printf("n");
printf("Vec to tri so du :n");
for (i=1;i<=n;i++)
printf("%.5fn",(a[i][n+1]));
mot:printf("n");
getch();
}
void doc()
{
a[1][1]=3*x[1]*x[1]-3*x[2]*x[4];
a[1][2]=-3*x[2]*x[2]-3*x[1]*x[4];
a[1][3]=0;
a[1][4]=-3*x[1]*x[2];
a[1][5]=x[1]*x[1]*x[1]-x[2]*x[2]*x[2]-3*x[1]*x[2]*x[4]-8;
a[2][1]=1;
a[2][2]=1;
a[2][3]=1;
a[2][4]=1;
a[2][5]=x[1]+x[2]+x[3]+x[4]-5;
a[3][1]=-x[1]/sqrt(25-x[1]*x[1]);
a[3][2]=0;
a[3][3]=8;
a[3][4]=0;
a[3][5]=sqrt(25-x[1]*x[1])+8*x[3]+4;
a[4][1]=2*x[2]*x[3];
a[4][2]=2*x[1]*x[3];
a[4][3]=2*x[1]*x[2];
a[4][4]=-1;
a[4][5]=2*x[1]*x[2]*x[3]-x[4]+8;
}